Consider the unit sphere $S=\{(x,y,z)\in {R^3}:x^2+y^2+z^2=1\}$ and the unit normal vector $\bar{n}$ at each point $(x,y,z)$ on $S$. Then the value of the surface integral $$\int\int_{S}\{(\frac{2x}{\pi}+\sin(y^2))x+(e^z-\frac{y}{\pi})y+(\frac{2z}{\pi}+\sin^2y)z\}d\sigma$$ is____?
Ok so this is what I tried to do:
Since this is a surface integral over a sphere, therefore I thought to convert it into a triple integral by using Gauss's Theorem, viz,
$$\int\int_{S}(f ~dy ~dz+g~dz~dx+h~dx~dy)=\int\int\int_{E}(f_x+g_y+h_z)~dx~dy~dz$$
So $f=(e^z-\frac{y}{\pi})y+(\frac{2z}{\pi}+\sin^2y)z$, $~g=0$ (since there is no term involving $~dz~dx$), and $h=(\frac{2x}{\pi}+\sin(y^2))x$. At this point I am stuck. This way doesnot provide any sensible conclusion.
Can anyone provide some other way? How to solve this problem? Thanks.
Recall
\begin{align} \iint_\Sigma \mathbf{F}\cdot\mathbf{n}\ dS = \iiint_D \operatorname{div}\mathbf{F}\ dV \end{align} Here $\mathbf{F} = (\frac{2x}{\pi}+\sin(y^2))\mathbf{i}+(e^z-\frac{y}{\pi})\mathbf{j}+(\frac{2z}{\pi}+\sin^2(y))\mathbf{k}$ where $\operatorname{div}\mathbf{F} = \frac{3}{\pi}$.